fscanf with pointer of strings *string[]

Hi there,

I am reading a file that contains both numbers and strings; by reading it sequentially, everything goes well when reading floats and assigning their values to an array previously declared, BUT, when I read strings and try to assign them to a pointer defined as

Code:

const char *problem[2];

and allocated as

Code:

for(i=0; i<2; ++i )
problem[i] = (char*) malloc(12 * sizeof(char));

I obtain a segnemtation fault at the moment of reading the values.

The error is avoided ONLY if I assing the string to simple strings instead of the elements of a array of strings.

Could anyone help?

Code:

//Declare:
char header[12];
const char *problem[2];

//Allocate:
for(i=0; i<2; ++i )
problem[i] = (char*) malloc(12 * sizeof(char));

//Read file:

       file_ID = fopen(input_file, "r");
       fscanf(file_ID, "%s %s %s\n", header, &problem[0], &problem[1]);
     

      printf(" Verify problem 0: %s\n", problem[0]);
      printf(" Verify problem 1: %s\n", problem[1]);

I get a segmentation fault when printing it to file (if I comment print I dont)

the file to be read is the following:

Code:

problem:	 problem_name   problem_type

Thank you very much in advance
All the best
S.M.

I solved it in the following way, although I would have liked for the pointers to be assgigned the strings directly through fscanf:

Code:

         fscanf(file_ID, "%s %s %s\n", header, prob, prob_type);
	
	strcpy(problem[0], prob);
	strcpy(problem[1], prob_type);

It now works correctly.
S.M.

This will work:

Code:

fscanf(file_ID, "%s %s %s\n", header, problem[0], problem[1]);

no &; problem[0] is a pointer (you are used to passing the address of an int variable)

Originally Posted by MK27

This will work:

Code:

fscanf(file_ID, "%s %s %s\n", header, problem[0], problem[1]);

no &; problem[0] is a pointer (you are used to passing the address of an int variable)

It indeed does, but why this though?

thank you very much for helping!

Originally Posted by simone.marras

It indeed does, but why this though?

Since problem is an array of two pointers to const char, problem[0] is a pointer to const char.

Thank you
S.

However, it is equally true if problem[0] were a pointer to a regular char; you do need the address of operator when passing a pointer to a scanf function. You only need the address of if you are passing a variable which is not a pointer, such as an int.

This can lead to confusion because you CAN use &charptr with scanf and it will work, but that is not the way you SHOULD have been doing it.

Originally Posted by MK27

This can lead to confusion because you CAN use &charptr with scanf

only because pointer to array and pointer to the first element of the array contain same address you can sometimes get through with this incorrect code.

when charptr is really a pointer its address is different from its cotents...

that's why the OP's code crashed